Question

The percentage of fe in an iron ii sample was determined by redox titration with cr2o7 calculate the molarity of the standard k2cr207

Answer 1

The molarity is 0.018M and the percentage is 8.46%.

Net ionic reaction is

$\mathrm{Cr} 2 \mathrm{O}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H} 2 \mathrm{O}$

$1 \mathrm{~mol}$ of $\mathrm{Cr} 2 \mathrm{O} 7(-2)$ reacts with $6 \mathrm{~mol}$of $\mathrm{Fe}(2+)$ to form $6 \mathrm{~mol}$of $\mathrm{Fe}(3+)$ and $2 \mathrm{~mol}$ of $\mathrm{Cr}(3+)$

Find molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

Molarity = moles of$\mathrm{K} 2 \mathrm{Cr} 207 /$ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7(\mathrm{~L}$ )

Moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ = grams of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ molar mass of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

$=1.2275 \mathrm{~g} / 294.19 \mathrm{~mol}=0.0042 \mathrm{~g} / \mathrm{mol}$

Molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7=$ moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$/ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$(L)

$=0.0042 \mathrm{~g} / \mathrm{mol} / 0.250 \mathrm{~L}=0.017 \mathrm{M}$

Find molarity of $\mathrm{Fe}$ (II)

Molarity of $\mathrm{Fe}(\mathrm{II})=6 \times$molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 \times$volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$volume of $\mathrm{Fe}$ (II)

$=6 \times 0.017 \mathrm{M} \times 0.03598 \mathrm{~L} / 0.200 \mathrm{~L}$

$=0.018 \mathrm{M}$

Find moles of $\mathrm{Fe}(\mathrm{II})$

Moles of $\mathrm{Fe}( II) =$molarity of $\mathrm{Fe}$ (II) $\times$ volume of $\mathrm{Fe}$ (II)

$=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}$

Find mass of Fe(II)

Mass of $\mathrm{Fe}( II )=$ moles of $\mathrm{Fe}( II$) $\times$ molar mass of $\mathrm{Fe}(\mathrm{II})$

$=0.004 \mathrm{~mol} \times 55.85 \mathrm{~g} / \mathrm{mol}$

$=0.205 \mathrm{~g}$

Find $\% \mathrm{Fe}(\mathrm{II})$ in unknown sample

$\begin{aligned}\% \mathrm{Fe} &=(\text { mass of } \mathrm{Fe} / \text { weight of unknown sample }) \times 100 \\&=(0.205 \mathrm{~g} / 2.4234 \mathrm{~g}) \times 100 \\&=8.46 \%\end{aligned}$

• Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution.

• concentration is the amount of a substance per defined space. Concentration usually is expressed in terms of mass per unit volume.

To know more about  MOLARITY    visit : brainly.com/question/8732513

#SPJ4